Monday, December 21, 2009

HERE IS THE RIDDLE: a line of 200 people waits to get into the theater with 200 seats. Each person has an assigned seat- except for the person at the front of the line, who is a VIP and can sit anywhere he wants. Any person with an assigned seat who finds that his or her seat is taken will pick another seat at random. What is the likelihood that the last person will get his assigned seat?

10 comments:

there is a 50% chance. the VIP will almost certainly sit in an assigned seat, and then everyone after him will sit in a random seat until someone sits in the one seat that was not assigned, at which point everyone will be able to sit in their assigned seat. the chances of the first person getting the unassigned seat are 1/200. the second person has a 1/199 odds, and so on. if you sum these odds you get 50 percent...

Assuming that the VIP favors each of the 200 seats equally, then the greatest likelihood that the last person sits in her assigned seat is 1/200. The simplest way for the last person to get her assigned seat is if the VIP selects the unassigned seat, which has a likelihood of 1/200. There are other ways in which the last person can get her assigned seat, but those ways have a lesser probability of occurring. For example, the VIP sits in person 2's seat, person 2 sits in person 3's seat, and person 3 sits in VIP's seat, which has a likelihood of 199/200 * 1/199 * 1/198 ~ 1/400 and everyone after person 3 gets his/her assigned seat. If the VIP chose person 2's seat and person 2 chose the VIP's seat, everyone afterwards gets his/her assigned seat and this case also has a probability of 199/200 * 1/199 = 1/200 of occurring.

0.5% chance. The VIP will almost certainly sit in a seat assigned to someone else which will in turn cause everyone else to be relocated. By the time the last person enters the seat left will almost definitely not be the one assigned to them.

There is a 50% chance that the last person will get his assigned seat?

Oh, you want to know why. ;-) Here's how I did it. I start with less seats.

2 Seats (The VIP and the Last Person): Obviously a 50% chance that the last person will get his assigned seat.

3 Seats (The VIP, Random Person, and the Last Person) 3 Possible Outcomesa.) The VIP sits in her seat, so a 100% chance for the Last Person to get his assigned seat. b.) The VIP sits in the Random Person seat. Now he's got a 50/50 chance of sitting in the Last Persons assigned seat. So 50% for him. c.) The VIP sits in Last Person’s assigned seat. 0% chance for him to get his seat.

So with 3 seats the chance is: 1/3*100% + 1/3*50% + 1/3*0% = 50%

4 seats (The VIP, Random Person 1, Random Person 2 and the Last Person)]: 4 Possible Outcomes a.) The VIP sits in her seat, so a 100% chance for the Last Person to get his assigned seat.b.) The VIP sits in Random Person 1’s seat. Now, he has to choose a seat at random, but if he sits in the VIP’s assigned seat they two of them have swapped, and everyone else is good. We have now got the same situation as we had above with three seats. One seat that is 'right' and two seats that are 'wrong'. We know from above that this situation results in there being a 50% chance for the Last Person to get his assigned seat. c.) The VIP sits in Random Person 2’s seat. Now Random Person 1 will go to his, and Random Person 2 has to make a 50/50 choice. This is the same as the 2 seat situation above and obviously there is a 50% chance for the Last Person to get his assigned seat.d.) The VIP sits in the Last Person’s seat. 0% chance for him to get his seat.

So with 4 seats: 1/4*100% + 1/4*50% + 1/4*50% + 1/4*0% = 50%.

As you can see any number of seats reduces to the same situation. There is always a 50/50 chance for the Last Person to get his assigned seat.